Prove that

Solution:

To Prove: $2 \cos ^{2} 15^{\circ}-1=\frac{\sqrt{3}}{2}$

Taking LHS,

$=2 \cos ^{2} 15^{\circ}-1 \ldots(\mathrm{i})$

We know that,

$1+\cos 2 x=2 \cos ^{2} x$

Here, $x=15^{\circ}$

So, eq. (i) become

$=\left[1+\cos 2\left(15^{\circ}\right)\right]-1$

$=1+\cos 30^{\circ}-1$

$=\cos 30^{\circ}\left[\because \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}\right]$

$=\frac{\sqrt{3}}{2}$

= RHS

∴ LHS = RHS

Hence Proved